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            <h1 id="seo-header">『算法-ACM竞赛-图论』一般带花树匹配</h1>
            
            
              <div class="markdown-body">
                
                <h1 id="『算法-ACM-竞赛-图论』一般带花树匹配"><a href="#『算法-ACM-竞赛-图论』一般带花树匹配" class="headerlink" title="『算法-ACM 竞赛-图论』一般带花树匹配"></a>『算法-ACM 竞赛-图论』一般带花树匹配</h1><h1 id="图论–一般带花树匹配"><a href="#图论–一般带花树匹配" class="headerlink" title="图论–一般带花树匹配"></a>图论–一般带花树匹配</h1><p>带花树就是说一个非二分图，图中带有奇环的图，我们不能在奇环中找增广路，因为会陷入死循环，我们可以将带花树的花（奇环）部分缩成点处理，剩下的图就是一个无奇环的图。我们再找增广路，而奇环中的的点我们可以随意分配，但是说起来简单，但是实现很难。经过前人的探索，还有这篇《Efficient Algorithms for Finding Maximal Matching in Graphs》论文，呃，然后后人就写出来模板，这就是一个模板题。</p>
<p>模板：</p>
<pre><code class="hljs">#include &lt;cstdio&gt;
#include &lt;cstring&gt;
#include &lt;iostream&gt;
#include &lt;queue&gt;
using namespace std;
const int N = 250;
// 并查集维护
int belong[N];
int findb(int x) &#123;
    return belong[x] == x ? x : belong[x] = findb(belong[x]);
&#125;
void unit(int a, int b) &#123;
    a = findb(a);
    b = findb(b);
    if (a != b) belong[a] = b;
&#125;

int n, match[N];
vector&lt;int&gt; e[N];
int Q[N], rear;
int _next[N], mark[N], vis[N];
// 朴素算法求某阶段中搜索树上两点x, y的最近公共祖先r
int LCA(int x, int y) &#123;
    static int t = 0; t++;
    while (true) &#123;
        if (x != -1) &#123;
            x = findb(x); // 点要对应到对应的花上去
            if (vis[x] == t)
                return x;
            vis[x] = t;
            if (match[x] != -1)
                x = _next[match[x]];
            else x = -1;
        &#125;
        swap(x, y);
    &#125;
&#125;

void group(int a, int p) &#123;
    while (a != p) &#123;
        int b = match[a], c = _next[b];

        // _next数组是用来标记花朵中的路径的，综合match数组来用，实际上形成了
        // 双向链表，如(x, y)是匹配的，_next[x]和_next[y]就可以指两个方向了。
        if (findb(c) != p) _next[c] = b;

        // 奇环中的点都有机会向环外找到匹配，所以都要标记成S型点加到队列中去，
        // 因环内的匹配数已饱和，因此这些点最多只允许匹配成功一个点，在aug中
        // 每次匹配到一个点就break终止了当前阶段的搜索，并且下阶段的标记是重
        // 新来过的，这样做就是为了保证这一点。
        if (mark[b] == 2) mark[Q[rear++] = b] = 1;
        if (mark[c] == 2) mark[Q[rear++] = c] = 1;

        unit(a, b); unit(b, c);
        a = c;
    &#125;
&#125;

// 增广
void aug(int s) &#123;
    for (int i = 0; i &lt; n; i++) // 每个阶段都要重新标记
        _next[i] = -1, belong[i] = i, mark[i] = 0, vis[i] = -1;
    mark[s] = 1;
    Q[0] = s; rear = 1;
    for (int front = 0; match[s] == -1 &amp;&amp; front &lt; rear; front++) &#123;
        int x = Q[front]; // 队列Q中的点都是S型的
        for (int i = 0; i &lt; (int)e[x].size(); i++) &#123;
            int y = e[x][i];
            if (match[x] == y) continue; // x与y已匹配，忽略
            if (findb(x) == findb(y)) continue; // x与y同在一朵花，忽略
            if (mark[y] == 2) continue; // y是T型点，忽略
            if (mark[y] == 1) &#123; // y是S型点，奇环缩点
                int r = LCA(x, y); // r为从i和j到s的路径上的第一个公共节点
                if (findb(x) != r) _next[x] = y; // r和x不在同一个花朵，_next标记花朵内路径
                if (findb(y) != r) _next[y] = x; // r和y不在同一个花朵，_next标记花朵内路径

                                                // 将整个r -- x - y --- r的奇环缩成点，r作为这个环的标记节点，相当于论文中的超级节点
                group(x, r); // 缩路径r --- x为点
                group(y, r); // 缩路径r --- y为点
            &#125;
            else if (match[y] == -1) &#123; // y自由，可以增广，R12规则处理
                _next[y] = x;
                for (int u = y; u != -1; ) &#123; // 交叉链取反
                    int v = _next[u];
                    int mv = match[v];
                    match[v] = u, match[u] = v;
                    u = mv;
                &#125;
                break; // 搜索成功，退出循环将进入下一阶段
            &#125;
            else &#123; // 当前搜索的交叉链+y+match[y]形成新的交叉链，将match[y]加入队列作为待搜节点
                _next[y] = x;
                mark[Q[rear++] = match[y]] = 1; // match[y]也是S型的
                mark[y] = 2; // y标记成T型
            &#125;
        &#125;
    &#125;
&#125;

bool g[N][N];
int main() &#123;
    scanf(&quot;%d&quot;, &amp;n);
    for (int i = 0; i &lt; n; i++)
        for (int j = 0; j &lt; n; j++) g[i][j] = false;

    // 建图，双向边
    int x, y; while (scanf(&quot;%d%d&quot;, &amp;x, &amp;y) != EOF) &#123;
        x--, y--;
        if (x != y &amp;&amp; !g[x][y])
            e[x].push_back(y), e[y].push_back(x);
        g[x][y] = g[y][x] = true;
    &#125;

    // 增广匹配
    for (int i = 0; i &lt; n; i++) match[i] = -1;
    for (int i = 0; i &lt; n; i++) if (match[i] == -1) aug(i);

    // 输出答案
    int tot = 0;
    for (int i = 0; i &lt; n; i++) if (match[i] != -1) tot++;
    printf(&quot;%d\n&quot;, tot);
    for (int i = 0; i &lt; n; i++) if (match[i] &gt; i)
        printf(&quot;%d %d\n&quot;, i + 1, match[i] + 1);
    return 0;
&#125;
</code></pre>

                
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